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
=290.7gmol−1
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.