A mixture of 1.65 times 10^{21} molecules of | vedclass

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(C) Let the molecular mass of $X$ be $m \ g/mol$.The number of moles of $X$ is $n_X = \frac{1.65 	imes 10^{21}}{6.022 	imes 10^{23}} \ mol$.The mass

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$42$

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(C) Let the molecular mass of $X$ be $m \ g/mol$.The number of moles of $X$ is $n_X = \frac{1.65 	imes 10^{21}}{6.022 	imes 10^{23}} \ mol$.The mass of $X$ is $m_X = n_X 	imes m = \frac{1.65 	imes 10^{21}}{6.022 	imes 10^{23}} 	imes m \ g$.The number of moles of $Y$ is $n_Y = \frac{1.85 	imes 10^{21}}{6.022 	imes 10^{23}} \ mol$.The mass of $Y$ is $m_Y = n_Y 	imes 187 = \frac{1.85 	imes 10^{21}}{6.022 	imes 10^{23}} 	imes 187 \ g$.Given that the total mass $m_X + m_Y = 0.688 \ g$.Substituting the values: $\frac{1.65 	imes 10^{21} 	imes m}{6.022 	imes 10^{23}} + \frac{1.85 	imes 10^{21} 	imes 187}{6.022 	imes 10^{23}} = 0.688$.Multiplying by $N_A = 6.022 	imes 10^{23}$:$(1.65 	imes 10^{21}) 	imes m + (1.85 	imes 10^{21}) 	imes 187 = 0.688 	imes 6.022 	imes 10^{23}$.$1.65 	imes m + 1.85 	imes 187 = 0.688 	imes 602.2$.$1.65 	imes m + 345.95 = 414.3136$.$1.65 	imes m = 68.3636$.$m = \frac{68.3636}{1.65} \approx 41.43 \approx 42 \ g/mol$.

$A$ mixture of $1.65 \times 10^{21}$ molecules of $X$ and $1.85 \times 10^{21}$ molecules of $Y$ weighs $0.688 \ g$. If the molecular mass of $Y$ is $187 \ g/mol$,what is the molecular mass of $X$?

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