The weight of sodium hydrogen photphate (Na_2HPO_4) required to make 22.50 L of

Question

The weight of sodium hydrogen photphate (Na_2HPO_4) required to make 22.50 L of soltion0.0832 M NA-2HPO_4, is: The volume of a solution of 0.200 m C_6H_12O_6, in water required to make 120 ml of 0 00555 glucose solution (standard glucose solution for blood tests) The morality (molar concentration) of a solution of Ba(OH)_2 is determined by titrating it with standard solution of 0.400 M HCl. The volume of the HCl solution required to titrate 20 ml of If Ba(OH)_2 solution was found to be 62.5 mL. The molarity of the Ba(OH)_2 solution is; The only base present in a sample of 6 25 g of a certain solid mixture is sodium hydroxide In onto to determine the percentage of NaOH in this sample, the sample is dissolved in water and with 0.800 M solution of H_2SO_4 The volume of the H_2SO_4 required to complete the titration is for be 32.0 mL. The percentage of NaOH in the solid sample is:

Explanation / Answer

9: Moles of Na2HPO4 present in 22.50L of 0.0832 M solution = MxV = 0.0832M x 22.50L = 1.872 mol

Molecular mass of Na2HPO4 = 142 g/mol

Hence mass of Na2HPO4 required = 1.872 mol x (142 g/mol) = 265.75 g

Hence f is correct

10: For the required solution of glucose, V1 = 320 mL, M1 = 0.00555 M

The concentration of glucose solution taken initially,M2 = 0.200 M

Now law of dilution

M1V1 = M2V2

=> 320 mLx0.00555M = 0.200MxV2

=> V2 = ( 320 mLx0.00555M) / 0.200M = 8.88 mL (answer)

11: Given molarity of HCl = 0.400 M

Volume of HCl = 62.5 mL

Hence mmol of HCl = MxV(ml) = 0.400Mx62.5 mL = 25 mmol

Volume of Ba(OH)2 = 20 mL

Let's the concentration of Ba(OH)2 solution = M

Hence mmol of Ba(OH)2 = 20M

The neutralization is

Ba(OH)2 + 2 HCl ----- > BaCl2 + 2 H2O

1 mol 2 mol 1 mol 2 mol

for the above balanced reaction, 1 mmol of Ba(OH)2 reacts with 2 mmol of HCl.

Hence the moles of HCl that will react with 20M mmol of Ba(OH)2

= 20M Ba(OH)2 x (2 mol HCl / 1 mol Ba(OH)2) = 40 M Ba(OH)2

Hence 40 M = 25

=> M = 25/40 = 0.625 M (g)

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