One mole of HCl is required to prepare 1 L of 1N solution.
Calculate number of moles (n) required to prepare 200/1000 L of 0.5N solution.
Find molecular mass M of HCl in g
mass of pure HCl required = nM g
Volume of given acid required
= nM ÷ (1.18*0.37) g÷(g/mL)
how many millimeter should be taken out to prepare 0.5 normal HCL solution of 200 ml. density=1.18, purity 37%
please response soon
2 answers
Here is slightly different format.
First determine the N of the HCl. The molar mass HCl is about 36.5 g/mol.
1.18g/mL x 1000 mL x 0.37 x 1/36.5 = ? and that is approx 12N
Then use the dilution formula of
mL1 x N1 = mL2 x N2
mL1 x 12 = 200 x 0.5
Solve for mL1.
First determine the N of the HCl. The molar mass HCl is about 36.5 g/mol.
1.18g/mL x 1000 mL x 0.37 x 1/36.5 = ? and that is approx 12N
Then use the dilution formula of
mL1 x N1 = mL2 x N2
mL1 x 12 = 200 x 0.5
Solve for mL1.
1 answer