The density of 'x' M solution ('x' molar) of NaOH is \(1.12 \, \text{g/mL}\). While in molality, the concentration of the solution is \(3 \, \text{m}\) (3 molal). Then x is:
Given: Molar mass of NaOH is \(40 \, \text{g/mol}\)
Given: Molar mass of NaOH is \(40 \, \text{g/mol}\)
- 3.5
- 3
- 3.8
- 2.8
The Correct Option is B
Approach Solution - 1
We know the relationship between molality and molarity:
\[\text{Molality} = \frac{1000 \times M}{1000 \times d - M \times (\text{Molar mass of solute})}\]
Substituting the values:
\[3 = \frac{1000 \times x}{1000 \times 1.12 - x \times 40}\]
Rearranging:
\[3 \times (1000 \times 1.12 - x \times 40) = 1000 \times x\]
Simplify:
\[3 \times 1120 - 120x = 1000x\]
\[3360 = 1120x\]
Solving:
\[x = 3\]
Thus, the molarity of the solution is 3.0 M.
Approach Solution -2
To determine the molarity (\(x\)) of the NaOH solution, we need to relate molarity and molality using the density of the solution. The given molality is \(3 \, \text{m}\), meaning there are 3 moles of NaOH per kilogram of the solvent (water), and the density of the solution is \(1.12 \, \text{g/mL}\).
- Calculate the mass of the solution:
- The density of the solution implies that \(1 \, \text{mL}\) of solution weighs \(1.12 \, \text{g}\).
- Consider a total solution of \(1000 \, \text{mL}\) to find its mass:
- Mass of solution = \(1.12 \, \text{g/mL} \times 1000 \, \text{mL} = 1120 \, \text{g}\).
- Determine the mass of NaOH in the solution:
- Molality (\(3 \, \text{m}\)) means 3 moles of NaOH per 1000 g of water.
- Mass of water = \(1120 \, \text{g} - \text{mass of NaOH}\).
- If \(m\) is the mass of NaOH, \(1000 \, \text{g} - m\) of water also needs to allow for 3 moles of NaOH.
- Relation for NaOH:
- Moles of NaOH = \(\frac{m}{40} = 3\) (since molar mass of NaOH is \(40 \, \text{g/mol}\)), leading to \(m = 3 \times 40 = 120 \, \text{g}\).
- Mass of the water is therefore \(1000 \, \text{g} - 120 \, \text{g} = 880 \, \text{g} = 0.88 \, \text{kg}\).
- Calculate the molarity (\(x\)):
- Moles of solute (NaOH) in \(1120 \, \text{g} \, \text{solution}\) = 3 moles (as calculated).
- Molarity \((x) = \frac{\text{moles of solute}}{\text{liters of solution}} = \frac{3}{1} = 3 \, \text{M}\).
Hence, the molarity of the given solution is \(x = 3\).
Top JEE Main Chemistry Questions
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)- At equilibrium the concentrations of and in a sealed vessel at . What will be for the reaction
- Find out the major products from the following reactions
Cobalt chloride when dissolved in water forms pink colored complex $X$ which has octahedral geometry. This solution on treating with cone $HCl$ forms deep blue complex, $\underline{Y}$ which has a $\underline{Z}$ geometry $X, Y$ and $Z$, respectively, are
- The correct order of atomic radii is:
Top JEE Main Solutions Questions
- What weight of glucose must be dissolved in 100 g of water to lower the vapour pressure by 0.20 mm Hg?
(Assume dilute solution is being formed)
Given : Vapour pressure of pure water is 54.2 mm Hg at room temperature. Molar mass of glucose is 180 g mol–1 . - Solubility of \(Ca_3(PO_4)_2\) in \(100\) \(mL\) of pure water is \(W\) gm. Find out \(K_{sp}\) of \(Ca_3(PO_4)_2\) is: (M: Molecular mass of \(Ca_3(PO_4)_2\))
- We are given with three NaCl samples and their Van't Hoff factors. Which is correct option ?
Sample Van't Haff Factor Sample - 1 (0.1 M) \(i_1\) Sample - 2 (0.01 M) \(i_2\) Sample - 3 (0.001 M) \(i_2\) - Ethylene glycol of \(x\) \(kg\) is mixed with \(18.6\) \(kg\) of solvent, \(24 °C\) depression in freezing point takes place. Calculate value of \(x\). (Given: \(K_1\) = \(1.6 °C/molal\) M.W. of ethylene glycol = \(62\) \(g/mol\))
- $\text{Volume of } 3 \, \text{M NaOH (formula weight 40 g mol}^{-1}\text{) which can be prepared from } 84 \, \text{g of NaOH is } \_\_\_\_\_ \times 10^{-1} \, \text{dm}^3.$
Top JEE Main Questions
- In a hydrogen like atom, when an electron jumps from the $M$ - shell to the $L$ - shell, the wavelength of emitted radiation is $\lambda$. If an electron jumps from $N$-shell to the $L$-shell, the wavelength of emitted radiation will be :
- Two masses $m$ and $\frac{m}{2}$ are connected at the two ends of a massless rigid rod of length $l$. The rod is suspended by a thin wire of torsional constant $k$ at the centre of mass of the rod-mass system(see figure). Because of torsional constant $k$, the restoring torque is $\tau =k\theta$ for angular displacement $\theta$. If the rod is rota ted by $\theta_0$ and released, the tension in it when it passes through its mean position will be:
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)- The value of is
- The number of terms of an is even. The sum of the odd terms is and of the even terms is . The last term exceeds the first by . Then the number of terms in the series is ______.