Question

Similar to the % labeling of oleum, a mixture of $H_{3}P{O}_4$ and $P_4{O}_{10}$ is labelled as $(100+x)\%$, where $x$ is the maximum $P_4{O}_{10}$ present in $100$ g mixture of $H_{3}P{O}_4$ and $P_4{O}_{10}$. If such a mixture is labelled as $127\%$, then the mass of $P_4{O}_{10}$ present in $100$ g of the mixture is:

• A. $71$ g
• B. $47$ g
• C. $83$ g
• D. $35$ g

Answer 1

The correct option is A $71$ g

Similar to the % labelling of oleum,
$P_4{O}_{10}+6H_{2}O\to 4H_{3}P{O}_4$

$127$% mixture here means that $27g$ of $H_{2}O$ reacts with $100g$ mixture to give $127g$ $H_{3}P{O}_4$

$P_4{O}_{10}+6H_{2}O\to 4H_{3}P{O}_4$

Moles of $H_{2}O$ is $\frac{27}{18}=1.5$ moles

Now, moles of $P_4{O}_{10}$ will be $\frac{1.5}{6}=$ $\frac{1}{4}$ moles of $P_4{O}_{10}$

$\frac{1}{4}\times$ wt of $P_4{O}_{10}=$ Mass of$P_4{O}_{10}$

$\frac{1}{4}\times 284=71$ g of $P_4{O}_{10}$