What is the percent cscl by mass in a 0.711 m aqueous cscl solution that has a density of 1.091 g/ml?

Question

Madyson Huber

Chemistry

25 April, 00:35

What is the percent cscl by mass in a 0.711 m aqueous cscl solution that has a density of 1.091 g/ml?

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Answers (1)

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Claire Barrett · Answer 1

Answer is: the percent caesium chloride by mass is 11%.

V (CsCl) = 1 L · 1000 mL/L = 1000 mL.

c (CsCl) = 0.711 M.

n (CsCl) = V (CsCl) · c (CsCl).

n (CsCl) = 1 L · 0.711 mol/L.

n (CsCl) = 0.711 mol.

m (CsCl) = n (CsCl) · M (CsCl).

m (CsCl) = 0.711 mol · 168.36 g/mol.

m (CsCl) = 119.7 g.

d (CsCl) = 1.091 g/mL.

mr (CsCl) = d (CsCl) · V (CsCl).

mr (CsCl) = 1091 g.

w/w = 119.7 g : 1091 g * 100%.

w/w = 11 %.