A student prepares 150.0 mL of 1.40 M HCl using 35.0 mL of a stock solution. What is the concentration of the stock solution?

In dilution of a certain solution, the product of the volume and molarity of the initial solution should be the same as that of the resulting diluted solution. Such that M1V1 = M2V2. Substituting the known values, (150 mL) (1.40M) = (35.0 mL) (x M). The value of x from the equation is 6.

The concentration of the stock solution used to prepare 150.0 mL of 1.40 M HCl is 6 M. To calculate this we will use the formula C1 * V1 = C2 * V2. For the stock solution: C1 = ? and V1 = 35.0 mL. For the prepared solution: C2 = 1.40 M and V2 = 150 mL. Now, substitute this in the formula: C1 * 35.0 mL = 1.40 M * 150 mL. C1 = 1.40 M * 150 mL / 35.0 mL = 6 M.