Question

200 ml of a solution contains 0.754 g of a sodium bicarbonate Na HCO3 what is the normality​

Answer 1

Explanation:

200 ml of a solution contains 0.754 g of a sodium bicarbonate Na HCO3 what is the normality

Answer 2

Explanation:

Molarity of N2CO3 solution is M1 = (W/GMW) * 1000/V = (2.65/106) * 1000/250 = 0.1 molar Now using dilution law Molarity of resultant solution (M2) is M1V1 = M2V2 M2 = M1V1 / V2 M2 = (0.1 * 10) / 1000 M2 = 0.001 molar Concentration of resultant solution is 0.001M