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Standard XII

Chemistry

Gay Lussac's Law, Avagadro's Law

Two oleum sam...

Question

Two oleum samples $A$ and $B$ have labeling $109$ and $118$ respectively. $100 g$ of $A$ is diluted to an a $1 L$ solution and $100 g$ of $B$ is diluted to a $1000 m L$ solution. If both these solutions are mixed, then normality of $H_{2} S O_{4}$ (approx.) in final solution is:

1.16

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2.04

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3.48

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4.12

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Solution

The correct option is A $2.04$
Sample A
m $= 100$ g
M $= 98$
No. of mole $= \frac{100}{98} = 1.02$
Sample B
m $= 100$ g
M $= 98$
No. of mole $= \frac{100}{98} = 1.02$
Molarity(A) $= \frac{1.02}{1} = 1.02$
Molarity(B) $= \frac{1.02}{1} = 1.02$

Normality=acidity $\times M o l a r i t y$
Acidity of $H_{2} S O_{4} = 2$
$N_{A} = 1.02 \times 2 = 2.04$
$N_{B} = 1.02 \times 2 = 2.04$
to find resultant normality
$\frac{N_{A} V_{A} + N_{B} V_{B}}{V_{A} + V_{B}} = N$
$V_{A} = V_{B} = 1$ L
$N = \frac{2.04 + 2.04}{2} = 2.04$

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Similar questions

Q. Oleum is considered as a solution of

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Q. Oleum is considered as a solution of

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H_{2} S O_{4}

will be formed when 100 g of oleum is diluted by 9 g of

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Two oleum samples A and B have labeling 109 and 118 respectively. 100 g of A is diluted to an a 1 L solution and 100 g of B is diluted to a 1000 mL solution. If both these solutions are mixed, then normality of H2SO4 (approx.) in final solution is:

The correct option is A 2.04Sample Am=100 gM=98No. of mole=10098=1.02Sample Bm=100 gM=98No. of mole=10098=1.02Molarity(A)=1.021=1.02Molarity(B)=1.021=1.02Normality=acidity×MolarityAcidity of H2SO4=2NA=1.02×2=2.04NB=1.02×2=2.04to find resultant normalityNAVA+NBVBVA+VB=NVA=VB=1LN=2.04+2.042=2.04

Two oleum samples $A$ and $B$ have labeling $109$ and $118$ respectively. $100 g$ of $A$ is diluted to an a $1 L$ solution and $100 g$ of $B$ is diluted to a $1000 m L$ solution. If both these solutions are mixed, then normality of $H_{2} S O_{4}$ (approx.) in final solution is: