請幫忙我解答這幾題普通化學題目

9) What is the molar concentration of a solution that is prepared by dissolving 4.79 g of benzoic acid (C6H5CO2H) in 100 ml of solution?
A) 5.84 M
B) 0.392 M
C) 0.0479 M
D) 0.514 M

Molecular mass of C6H5CO2H = 12x7 + 1x6 + 16x2 = 122
No. of moles of C6H5CO2H = 4.79/122 = 0.0392 mol
Volume of the solution = 100 ml = 0.1 L
Concentration of the solution = 0.0392/0.1 = 0.392 M


11) What mass of KMnO4 should be used to prepare 250 ml of a 2.68 × 10-3 M solution?
A) 0.106 g
B) 0.424 g
C) 0.670 g
D) 1.70 g

No. of moles of KMnO4 = (2.68 x 10^-3) x (250/1000) = 0.00067 mol
Formula mass of KMnO4 = 39 + 55 + 16x4 = 158
Mass of KMnO4 = 0.00067 x 158 = 0.106 g


13) What is the concentration of a solution prepared by taking 25.0 ml of 4.63 × 10-2 M glucose (C6H12O6) and diluting it to 2.00 liters?
A) 2.32 × 10-3M
B) 5.79 × 10-4M
C) 3.70 M
D) 0.579 M

No. of moles of C6H12O6 = (4.63 x 10^-2) x (25/1000) = 1.1575 x 10^-3 mol
Concentration of the diluted solution = (1.1575 x 10^-3)/2 = 5.79 x 10^-4 M

2010-03-30 12:48:08 補充：
9) 把 4.79 g 的苯酸（C6H5CO2H）溶於 100 ml 溶液，溶液的摩爾濃度是多少？

11) 製備 250 ml 的 2.68 x 10^-3 M 溶液，應使用多少質量的 KMnO4？

12) 把 25.0 ml 的 4.63 × 10-2 M 葡萄糖（C6H12O6）稀釋成 2 升，所製備的溶液的濃度是多少？