Of course 0.100 mol of AgCl.
2 m is the molarity of a solution that has 6 mol of CaCl2 in 3 km of water.
4 mol/0.800 kg
0.67
(.05)X(grams of total solution) = grams of acetic acid (grams of acetic acid)/ (mol. wt. of acetic acid(=60g/mol)) = mol. acetic acid (mol. acetic acid)/ (Liters of total solution) = molarity(M)
Of course 0.100 mol of AgCl.
.15
2 m is the molarity of a solution that has 6 mol of CaCl2 in 3 km of water.
4 mol/0.800 kg
0,4 mol NaCl is 23,376 g.2,85 mol NaCl is 166,554 g to 1L.140 mL solution NaCl 2,85 M contain 0,4 mol NaCl.
20.2 g of CuCl2 = .1502 mol CuCl2 M=mol/L M=.1502 mol/L
4 mol/0.800 kg
0.751 mol/0.951 L = 0.790 mol/liter = 0.790 M
The molarity is 2 mol/L.
0.67
molarity = moles/litre [solution] = 12 mol/6 L [solution] = 2 mol/L = 2 M
.0253M