Q16 / 120 · VITEEE 2008 · Chemistry

The number of formula units of calcium flouride $\mathrm{CaF}_{2}$ present in $146.4 \mathrm{~g}$ of $\mathrm{CaF}_{2}$ are (molar mass of $\mathrm{CaF}_{2}$ is $78.08 \mathrm{~g} / \mathrm{mol}$ )

  • Correct answer
    A
    $1.129 \times 10^{24} \mathrm{CaF}_{2}$
  • B
    $1.146 \times 10^{24} \mathrm{CaF}_{2}$
  • C
    $7.808 \times 10^{24} \mathrm{CaF}_{2}$
  • D
    $1.877 \times 10^{24} \mathrm{CaF}_{2}$

Solution

Formula unit $=$ no. of molecules of $\mathrm{CaF}_{2}$. $\begin{array}{l} \text { Moles }=\frac{\text { massin } \mathrm{gm}}{\text { molar mass }}=\frac{146.4 \mathrm{gm}}{78.08 \mathrm{gm}}=1.875 \\ \begin{aligned} \text { Molecules } &=\text { Mole } \times 6.022 \times 10^{23} \\ &=1.875 \times 6.022 \times 10^{23} \\ &=1.129 \times 10^{24} \text { units of } \mathrm{CaF}_{2} \end{aligned} \end{array}$

Hence, the correct answer is option: 'A'.

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