What is the concentration (M) of a NaCl solution prepared by dissolving 5.3 g of NaCl in sufficient water to give 115 mL of solution?

Concentration = Mass of the solute/Mass of the solution= 45/145 = 0.310.31*60 = 18.6g/mol

Molar concentration of nitrate ion=0.167M

Molar mass of NaCl= 58.4414.6/58.44=0.25mol0.25/0.1= 2.5mol/L

percent v/v is (mL ethanol/mL solution)×100.So you have 25.0 mL ethanol/250 mL(25/250)×100 = 10% v/v.

250 ml × 0.4 = 100ml= 100/2 = 50mlMolar mass of sodium hydrogen carbonate = 84.007Molarity = 111/84.007 = 1.32 mol1.32/20= 0.066mol/L

Answer: n(CaCl2) = m(CaCl2) / M(CaCl2) = 16.7 / 111 = 0.15 c(CaCl¬2) = n(CaCl2) / V(solution) = 0.15 / 0.4 = 0.375 M

The solution is 74.7% pure.

w=(m(salt))/m(solution) =(m(salt))/(m(water)+m(salt))×100%=(5.0 g)/(550 g+5.0g)×100%=0.90% Answer: 0.90%.

MOLAR MASS OF CUSO4.5H2O = 249.5 gm /mol ..given mass of CUSO4.5H2O = 0.10 g .So , moles of CUSO4.5H2O =( 0.10 / 249.5 ) mole .volume of water =...

273.5 ml