Question

(b) 500. ml of 0.500 m ba(no3)2
(c) 125 ml of 0.100 m sbcl3
(d) 250. ml of 0.100 m sbcl3 contained in 350.0 ml of 0.250 m alcl3?

1. how many moles of alcl3 can be made from 100.0 g of hcl?
2. what volume of 2.40 m hcl are contained in 55.0 ml of 1.30 × 10^-3 m sr(no3)2?
3. how many moles of sr(no3)2 naf contains 0.15 g of naf?
4. what volume of 2.8 × 10^-2 m is 1.000 kg/l. what is the molar concentration of h2o in pure water
5. the density of water at 4°c is 1.000 kg/l. what is the molar concentration of h2o in pure water

4°c? (hint: how many moles of h2o are contained in 1 l?)
66 the density of acetic acid, ch3cooh(l), is 1049 g/l. what is the molarity of pure acetic acid?

Explanation:

Problem 61

Step1: Calculate moles of HCl

Molar mass of HCl = 1.008 + 35.45 = 36.458 g/mol. Moles = mass/molar mass:
$n = \frac{100.0\ \text{g}}{36.458\ \text{g/mol}} \approx 2.743\ \text{mol}$

Step2: Find volume of 2.40 M HCl

Volume = moles/molarity:
$V = \frac{2.743\ \text{mol}}{2.40\ \text{mol/L}} \approx 1.14\ \text{L}$

Step1: Calculate moles of Sr(NO₃)₂

Moles = molarity × volume (in L):
$n = 1.30 \times 10^{-3}\ \text{mol/L} \times 0.0550\ \text{L} = 7.15 \times 10^{-5}\ \text{mol}$

Step1: Calculate moles of NaF

Molar mass of NaF = 22.99 + 19.00 = 41.99 g/mol. Moles = mass/molar mass:
$n = \frac{0.15\ \text{g}}{41.99\ \text{g/mol}} \approx 0.00357\ \text{mol}$

Step2: Find volume of NaF solution

Volume = moles/molarity:
$V = \frac{0.00357\ \text{mol}}{2.8 \times 10^{-2}\ \text{mol/L}} \approx 0.13\ \text{L}$

Answer:

1.14 L

Problem 62