0,35 kmol/m³ = 0,35 mol/dm³ = 0,35 mol/L
175 mL = 0,175 L
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C = n/V
n = 0,35×0,175
n = 0,06125 mol
mCa(NO₃)₂: 40+(14×2)+(16×6) = 164 g/mol
1 mol --------- 164g
0,06125 ---- X
X = 10,045g
To prepare 175 mL of 0,35M solution, add 10,045g of calcium nitrate and add water to a volume of 175ml.
2x 6.022x10^23= 1.204x10^24
