(C) Total mass of $H_2SO_4 = (100 \times 0.98) + (100 \times 0.49) = 98 + 49 = 147 \ g$. Total mass of solution $= 100 + 100 = 200 \ g$. Total mass of

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(C) Total mass of $H_2SO_4 = (100 \times 0.98) + (100 \times 0.49) = 98 + 49 = 147 \ g$. Total mass of solution $= 100 + 100 = 200 \ g$. Total mass of $H_2O = 200 - 147 = 53 \ g$. Moles of $H_2SO_4 = \frac{147}{98} = 1.5 \ mol$. Moles of $H_2O = \frac{53}{18} \approx 2.944 \ mol$. Mole fraction of $H_2SO_4 = \frac{n_{H_2SO_4}}{n_{H_2SO_4} + n_{H_2O}} = \frac{1.5}{1.5 + 2.944} = \frac{1.5}{4.444} \approx 0.337$.

Class 12 Chemistry Solutions Method of expressing concentration of solution

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At $T \ (K)$, $100 \ g$ of $98\%$ $H_2SO_4$ $(w/w)$ aqueous solution is mixed with $100 \ g$ of $49\%$ $H_2SO_4$ $(w/w)$ aqueous solution. What is the mole fraction of $H_2SO_4$ in the resultant solution?
(Given: Atomic mass $H = 1 \ u; S = 32 \ u; O = 16 \ u$)
(Assume that temperature after mixing remains constant)

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A
$0.9$
B
$0.1$
C
$0.337$
D
$0.663$

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Method of expressing concentration of solution

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Match List-$I$ with List-$II$.

List-$I$ (Example) List-$II$ (Type of Solution)

$A$. Humidity $I$. Solid in solid

$B$. Alloys $II$. Liquid in gas

$C$. Amalgams $III$. Solid in gas

$D$. Smoke $IV$. Liquid in solid

Choose the correct answer from the options given below:

List-$I$ (Example)	List-$II$ (Type of Solution)
$A$. Humidity	$I$. Solid in solid
$B$. Alloys	$II$. Liquid in gas
$C$. Amalgams	$III$. Solid in gas
$D$. Smoke	$IV$. Liquid in solid

DifficultNEET 2025

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$1 \ g$ of a solute is added to $18 \ g$ of water. In this aqueous solution,what is the mass per cent of the solute (in $\%$)?

EasyAP EAMCET 2020

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If $1 \ M$ and $2.5 \ L$ $NaOH$ solution is mixed with another $0.5 \ M$ and $3 \ L$ $NaOH$ solution,then the molarity of the resultant solution will be $........ \ M$.

MediumAIPMT 2002

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If the mole fraction of water in a solution of $H_2SO_4$ is $0.85$,then the molality of the solution is:

Medium

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The density of a solution prepared by dissolving $120 \ g$ of urea (mol. mass $= 60 \ u$) in $1000 \ g$ of water is $1.15 \ g/mL$. The molarity of this solution is : ............ $M$

DifficultAIEEE 2012

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Match List-$I$ with List-$II$. List-$I$ (Example) List-$II$ (Type of Solution) $A$. Humidity $I$. Solid in solid $B$. Alloys $II$. Liquid in gas $C$. Amalgams $III$. Solid in gas $D$. Smoke $IV$. Liquid in solid Choose the correct answer from the options given below:

$1 \ g$ of a solute is added to $18 \ g$ of water. In this aqueous solution,what is the mass per cent of the solute (in $\%$)?

If $1 \ M$ and $2.5 \ L$ $NaOH$ solution is mixed with another $0.5 \ M$ and $3 \ L$ $NaOH$ solution,then the molarity of the resultant solution will be $........ \ M$.

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The density of a solution prepared by dissolving $120 \ g$ of urea (mol. mass $= 60 \ u$) in $1000 \ g$ of water is $1.15 \ g/mL$. The molarity of this solution is : ............ $M$

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Match List-$I$ with List-$II$.

List-$I$ (Example) List-$II$ (Type of Solution)

$A$. Humidity $I$. Solid in solid

$B$. Alloys $II$. Liquid in gas

$C$. Amalgams $III$. Solid in gas

$D$. Smoke $IV$. Liquid in solid

Choose the correct answer from the options given below: