Question

what mass of pbcl₂ contains 1.0 × 10²⁴ atoms of cl?
a) 0.83 g b) 130 g c) 84 g
d) 230 g e) 250 g

Explanation:

Step1: Find moles of Cl atoms

Use Avogadro's number ($N_A = 6.022\times10^{23}\ \text{atoms/mol}$). Moles of Cl, $n_{\text{Cl}}=\frac{N}{N_A}=\frac{1.0\times10^{24}\ \text{atoms}}{6.022\times10^{23}\ \text{atoms/mol}}\approx1.66\ \text{mol}$.

Step2: Relate moles of Cl to $PbCl_2$

In $PbCl_2$, 1 mole of $PbCl_2$ contains 2 moles of Cl. So moles of $PbCl_2$, $n_{\text{PbCl}_2}=\frac{n_{\text{Cl}}}{2}=\frac{1.66\ \text{mol}}{2}\approx0.83\ \text{mol}$.

Step3: Calculate molar mass of $PbCl_2$

Molar mass of Pb = 207.2 g/mol, Cl = 35.45 g/mol. Molar mass of $PbCl_2$: $M = 207.2 + 2\times35.45 = 207.2 + 70.9 = 278.1\ \text{g/mol}$.

Step4: Find mass of $PbCl_2$

Use $m = n\times M$. $m = 0.83\ \text{mol}\times278.1\ \text{g/mol}\approx230\ \text{g}$? Wait, no—wait, recalculate step2: Wait, $n_{\text{Cl}}=\frac{1.0\times10^{24}}{6.022\times10^{23}}\approx1.66\ \text{mol}$. Then $n_{\text{PbCl}_2}=\frac{1.66}{2}=0.83\ \text{mol}$. Then $m = 0.83\ \text{mol}\times278.1\ \text{g/mol}\approx230\ \text{g}$? Wait, but let's check again. Wait, maybe I made a mistake. Wait, 1.0e24 / 6.022e23 is ~1.66 mol Cl. Then $PbCl_2$ is 1:2, so 0.83 mol $PbCl_2$. Molar mass of $PbCl_2$: 207.2 + 235.45 = 207.2 + 70.9 = 278.1 g/mol. Then 0.83 278.1 ≈ 230 g. So the answer should be d) 230 g.

Answer:

d) 230 g