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Question

Calculate the molarity of each of the following solutions:

A: 30 g of Co(NO3)2.6H2O in 4.3L of solution.

B: 30mL of 0.5 M H2SO4 diluted to 500 mL.

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Solution

A:

Given:
Mass of solute (w2)=30 g
Volume of solution =4.3 L
Molar mass of given solute CO(NO3)2.6H2O
=58.7+2[14+(16×3)]+6(2+16)
=58.7+(2×62)+(6×18)
=58.1+124+108
=290.7gmol1

Moles of CO(NO3)2.6H2O

Number of moles of CO(NO3)2.6H2O=(Msss of cobalt nitrate)(Molar mass of cobalt nitrate)

Number of moles of CO(NO3)2.6H2O=30290.7=0.103mol

Molarity of CO(NO3)2.6H2O

Molarity (M)=moles of soluteVolume of solution (L)

Molarity (M)=0.1034.3=0.0239M

Final Answer:

The molarity of
30g of Co(NO3)2.6H2O in 4.3L of solution is 0.0239M.

B:

Given:

30mL of 0.5 M H2SO4 dilutes to 500 mL

Molarity of Final mixture

For dilution from volume V1 to V2

M1V1=M2V2

putting the values,
V1=30 mL
M1=0.5 M
V2=500 mL

We get,

M2=30×0.5500=0.03M
Thus, final molarity =0.03M

The molarity of 30 mL of 0.5 M H2SO4 diluted to 500 mL is 0.03M.


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