calculate the number of milliliters of an aqueous .300M zinc nitrate solution required to provide 21.0 grams of zinc nitrate

Question

abmon98 · Answer

Number of Moles=Mass(g)/Molar Mass(g/mol)
Look out for the atomic weight in your periodic table for each element present in Zinc nitrate compound. Zn(NO3)2
Zn:65.4 N:14 O:16
21.0/65.4+(14*2)+(16*6)=21.0/189.4=0.110 moles
Number of Moles=Concentration(mol/dm^3)*Volume
0.110/0.300=Volume
Volume=0.369 dm^3 
To change from dm^3 to milliliters we multiply by 1000
369 milliliters

anonymous · Answer

How did you get .369? I'm getting .366