Question

A mixture of $3\times {10}^{21}$ molecules of $X$ and $4.5\times {10}^{21}$ molecules of $Y$ weigh $1.375g$. If the molecular weight of $X$ is $50g$, then what is the molecular weight of $Y$?

• A. $100$
• B. $150$
• C. $250$
• D. $300$

Answer 1

The correct option is B $150$

Given,

Number of molecules of $X=3\times {10}^{21}$

Number of molecules of $Y=4.5\times {10}^{25}$

Molecular weight of $X=50$

Total mass of mixture $=1g$

Formula used :

$Massofmixture=[\frac{NumberofmoleculesofX}{Avograd{o}^{\prime }sNumber}\times MolarMassofX]$

$+[\frac{NumberofmoleculesofY}{Avograd{o}^{\prime }sNumber}\times MolarMassofY]$

Now, putting all the given values in this formula, we get the molecular mass of $Y$.

$1=[\frac{3\times {10}^{21}}{6.023\times {10}^{23}}\times 50]+[\frac{4.5\times {10}^{25}}{6.023\times {10}^{23}}\times Y]$

$\Rightarrow 1=\frac{150\times {10}^{21}+4.5Y\times {10}^{25}}{6.023\times {10}^{23}}$

$\Rightarrow 6.023\times {10}^{23}={10}^{21}(150+4.5Y\times {10}^4)$

$\Rightarrow \frac{6.023\times {10}^{23}}{{10}^{21}}=150+4500Y$

$\Rightarrow 602.3=150+4500Y$

$\Rightarrow 4500Y=602.3-150$

$\Rightarrow Y=\frac{450.3}{4500}$

$Y=150$

Therefore, on solving, we get $Y=150$

Therefore, the molecular weight of $Y$ is $150g$.

Hence option (B) is correct.