Question

How many molecules are in 85 liters of AgNO3?
How many cadmium atoms are there in 6.57x 10 to the 23rd power moles?
Please help with these two question need for a test.

Answer 1

Answer on Question #40831 - Chemistry – Inorganic Chemistry

Question

How many molecules are in 85 liters of $\mathrm{AgNO}_3$?

How many cadmium atoms are there in $6.57 \times 10$ to the 23rd power moles?

Please help with these two questions need for a test.

Answer:

First question:

$\mathrm{AgNO}_3$ is a solid, so there is no sense to measure its volume. For solids the main criteria is their mass. It looks like there is a mistake in first question. Two variants are possible: 85 grams of $\mathrm{AgNO}_3$ or 85 liters of solution of $\mathrm{AgNO}_3$. As we don't have a concentration value in this question, we assume that the question is "How many molecules are in 85 grams of $\mathrm{AgNO}_3$?"

Number of moles of $\mathrm{AgNO}_3$ equals:

$$n \left(A g N O _ {3}\right) = \frac {m}{M}$$

m – Mass of $\mathrm{AgNO}_3$, m = 85 g.

M – Molar mass of $\mathrm{AgNO}_3$, g/mol:

$$M \left(A g N O _ {3}\right) = M (A g) + M (N) + 3 M (O) = 108 + 14 + 3 \cdot 16 = 170 \, g/mol$$

Then number of moles in 85 g of substance $\mathrm{AgNO}_3$ equals:

$$n \left(A g N O _ {3}\right) = \frac {85}{170} = 0.5 \, mol$$

Number of molecules of $\mathrm{AgNO}_3$ equals:

$$N = n \left(A g N O _ {3}\right) \cdot N _ {A}$$

$N_A$ – the Avogadro constant, $N_A = 6.022 \cdot 10^{23}$.

$$N = 0.5 \cdot 6.022 \cdot 10^{23} = 3.011 \cdot 10^{23}$$

Second question:

Number of cadmium atoms equals:

$$N = n (C d) \cdot N _ {A}$$

$n(\mathrm{Cd})$ – number of moles of cadmium, $n(\mathrm{Cd}) = 6.57 \cdot 10^{23} \, \mathrm{mol}$.

$N_A$ – the Avogadro constant, $N_A = 6.022 \cdot 10^{23}$.

$$N = 6.57 \cdot 10^{23} \cdot 6.022 \cdot 10^{23} = 3.956 \cdot 10^{47}$$

Answer: $3.011 \cdot 10^{23}$ molecules of $\mathrm{AgNO}_3$; $3.956 \cdot 10^{47}$ atoms of Cd.