A solution containing 306.5 g of Mg(NO 3 ) 2 per liter has a density of 1.114 g/

Question

A solution containing 306.5 g of Mg(NO3)2 per liter has a density of 1.114 g/mL. The molarity of the solution is:

A)

2.066 M

B)

1.855 M

C)

6.199 M

D)

2.302 M

E)

none of these

What is the molality of a solution of 30.1 g of propanol (CH3CH2CH2OH) in 152 mL water, if the density of water is 1.00 g/mL?

A)

3.29 m

B)

0.00330 m

C)

0.303 m

D)

0.501 m

E)

5.01 m

A solution containing 292 g of Mg(NO3)2 per liter has a density of 1.108 g/mL. The molality of the solution is:

A)

2.00 m

B)

2.41 m

C)

1.77 m

D)

6.39 m

E)

none of these

A)

2.066 M

B)

1.855 M

C)

6.199 M

D)

2.302 M

E)

none of these

Explanation / Answer

1)

Molar mass of Mg(NO3)2,
MM = 1*MM(Mg) + 2*MM(N) + 6*MM(O)
= 1*24.31 + 2*14.01 + 6*16.0
= 148.33 g/mol


mass(Mg(NO3)2)= 306.5 g

use:
number of mol of Mg(NO3)2,
n = mass of Mg(NO3)2/molar mass of Mg(NO3)2
=(306.5 g)/(148.33 g/mol)
= 2.066 mol
volume , V = 1 L


use:
Molarity,
M = number of mol / volume in L
= 2.066/1
= 2.066 M
Answer: A

2)

Molar mass of C3H8O,
MM = 3*MM(C) + 8*MM(H) + 1*MM(O)
= 3*12.01 + 8*1.008 + 1*16.0
= 60.094 g/mol


mass(C3H8O)= 30.1 g

use:
number of mol of C3H8O,
n = mass of C3H8O/molar mass of C3H8O
=(30.1 g)/(60.094 g/mol)
= 0.5009 mol

m(solvent)= density * volume
= 1.00 g/mL * 152 mL
= 152 g
= 0.152 kg

use:
Molality,
m = number of mol / mass of solvent in Kg
=(0.5009 mol)/(0.152 Kg)
= 3.29 molal
Answer: 3.29 molal

3)

Molar mass of Mg(NO3)2,
MM = 1*MM(Mg) + 2*MM(N) + 6*MM(O)
= 1*24.31 + 2*14.01 + 6*16.0
= 148.33 g/mol


mass(Mg(NO3)2)= 292 g

use:
number of mol of Mg(NO3)2,
n = mass of Mg(NO3)2/molar mass of Mg(NO3)2
=(292.0 g)/(148.33 g/mol)
= 1.969 mol

mass of solution = density * volume
= 1.108 g/ml * 1000 mL
= 1108 g
m(solvent)= 1108 g - 292 g
= 816 g
= 0.816 kg

use:
Molality,
m = number of mol / mass of solvent in Kg
=(1.969 mol)/(0.816 Kg)
= 2.41 molal
Answer: 2.41 molal

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