Question

match the questions to the correct answers.
column a

1. how many particles are in 14.9 liters of clf₃ at stp?
2. how many grams are in 11.6 liters of sbh₃ at stp?
3. how many grams are in 1.11 x 10²³ particles of sn(clo₄)₄?
4. how many formula units are in 633 grams of nahso₄?

column b
a. 3.91x10¹⁹
b. 3.17x10²⁴
c. 6.89
d. 95.2
e. 64.6
f. 1450
g. 4.00x10²³

Explanation:

Step1: Recall molar volume at STP

At STP (Standard Temperature and Pressure), 1 mole of any gas occupies 22.4 L.

Step2: Calculate moles of $ClF_3$

For the first question about $ClF_3$, $n=\frac{V}{V_m}=\frac{14.9\ L}{22.4\ L/mol}\approx0.665\ mol$. Using Avogadro's number $N = n\times N_A$, where $N_A = 6.022\times 10^{23}\ mol^{- 1}$, $N=0.665\ mol\times6.022\times 10^{23}\ mol^{-1}=3.91\times 10^{23}$. So 1 - a.

Step3: Calculate moles of $SbH_3$

For $SbH_3$, $n=\frac{V}{V_m}=\frac{11.6\ L}{22.4\ L/mol}\approx0.518\ mol$. The molar - mass of $SbH_3$ is $M = 121.76 + 3\times1.01=124.79\ g/mol$. The mass $m=n\times M = 0.518\ mol\times124.79\ g/mol\approx64.6\ g$. So 2 - e.

Step4: Calculate moles of $Sn(ClO_4)_4$

For $Sn(ClO_4)_4$, $n=\frac{N}{N_A}=\frac{1.11\times 10^{23}}{6.022\times 10^{23}\ mol^{-1}}\approx0.184\ mol$. The molar - mass of $Sn(ClO_4)_4$ is $M = 118.71+4\times(35.45 + 4\times16)=118.71+4\times(35.45 + 64)=118.71+4\times99.45=118.71 + 397.8 = 516.51\ g/mol$. The mass $m=n\times M=0.184\ mol\times516.51\ g/mol\approx95.2\ g$. So 3 - d.

Step5: Calculate moles of $NaHSO_4$

The molar - mass of $NaHSO_4$ is $M = 22.99+1.01 + 32.07+4\times16=120.07\ g/mol$. $n=\frac{m}{M}=\frac{633\ g}{120.07\ g/mol}\approx5.27\ mol$. The number of formula units $N = n\times N_A=5.27\ mol\times6.022\times 10^{23}\ mol^{-1}=3.17\times 10^{24}$. So 4 - b.

Answer:

1. a. $3.91\times 10^{23}$
2. e. $64.6$
3. d. $95.2$
4. b. $3.17\times 10^{24}$