A solution is prepared by adding 1 mole ethyl alcohol in 9 mole water. The mass percent of solute in the solution is ___________ (Integer Answer)
(Given : Molar mass in g mol$^{-1}$ Ethyl alcohol : 46, water : 18)
(Given : Molar mass in g mol$^{-1}$ Ethyl alcohol : 46, water : 18)
Correct Answer: 22
Approach Solution - 1
We are given the mass of ethyl alcohol and water, and we need to calculate the percentage by mass of ethyl alcohol in the solution:
Mass of ethyl alcohol = 1 mole × MM
$$ \Rightarrow 46 \, \text{g} $$Mass of water = 9 moles × MM
$$ \Rightarrow 162 \, \text{g} $$The percentage by mass of ethyl alcohol is given by:
$$ \% \, \text{by mass of ethyl alcohol} = \frac{46}{162 + 46} \times 100 $$ $$ \Rightarrow 22\% \, (\text{approx.}) $$Thus, the percentage by mass of ethyl alcohol is approximately 22%.
Approach Solution -2
Step 1: Calculate the mass of ethyl alcohol and water
\[ \text{Mass of ethyl alcohol} = 1 \, \text{mole} \times 46 \, \text{g mol}^{-1} = 46 \, \text{g}. \]
\[ \text{Mass of water} = 9 \, \text{mole} \times 18 \, \text{g mol}^{-1} = 162 \, \text{g}. \]
Step 2: Calculate total mass of solution
\[ \text{Total mass of solution} = \text{Mass of ethyl alcohol} + \text{Mass of water} = 46 + 162 = 208 \, \text{g}. \]
Step 3: Calculate mass percent of ethyl alcohol
\[ \text{Mass percent of ethyl alcohol} = \frac{\text{Mass of ethyl alcohol}}{\text{Total mass of solution}} \times 100. \]
\[ \text{Mass percent} = \frac{46}{208} \times 100 = 22.11\%. \]
Final Answer: 22
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
- 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 ______.
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,