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svetlana [45]
3 years ago
6

How do you prepare 250.00 mL of 3.00 M HCl solution starting with a concentrated HCl solution which is 36.0% HCl by mass and has

a density of 1.18 g/mL?
Chemistry
1 answer:
gogolik [260]3 years ago
6 0

Answer: 64.1 ml of concentrated HCl is taken and (250-64.4)= 185.6 ml f water is added to make 250.00 mL of 3.00 M HCl .

Explanation:

Given : 36 g of HCl is dissolved in 100 g of solution.

Density of solution = 1.18 g/ml

Thus volume of solution = \frac{mass}{density}=\frac{100}{1.18}=84.7ml

Molarity=\frac{n\times 1000}{V_s}

where,

n= moles of solute

Moles=\frac{\text{Given mass}}{\text{Molar mass}}=\frac{36g}{36.5g/mol}=0.99moles  

V_s = volume of solution

Molarity=\frac{0.99\times 1000}{84.7}=11.7M

According to the dilution law,

M_1V_1=M_2V_2

where,

M_1 = molarity of stock HCl solution = 11.7M

V_1 = volume of stock HCl solution = V ml

M_2 = molarity of diluted HCl solution = 3.00 M

V_2 = volume of diluted HCl solution = 250.0 ml

11.7M\times V=3.00\times 250.0

V=64.1ml

Thus 64.1 ml of stock HCl is taken and (250-64.4)= 185.6 ml of water is added to make 250.00 mL of 3.00 M HCl .

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