chemistry, stoichiometry and concentrations/ molar mass etc

Question

A) Formalin is an aqueous solution that is made from dissolving formaldehyde gas, HCHO (g), in water. A saturated formalin solution has a concentration of about 30.92% (m/v). This concentration is used to preserve biological specimens. Calculate the molar concentration of 20.92% (m/v) formalin.

B) excess sodium phosphate is mixed with 673.5 mL of strontium chloride at a concentration of 0.471 mol/L. If the reaction has a 49.33% yield, what is the mass of the dry precipitate actually collected?

C) what volume (ml) of 0.414 mol/L sodium hydroxide, NaOH (aq), is needed to completely neutralize 57.2 mL of 0.170 mol/L hydrochloric acid. HCI (aq)?

Answer 1

Formalin has the chemical formula H₂C=O and thus has a molar mass of 30.0 g/mol.

You asked for the molar concentration of 20.92% (m/v) solution, but I think you meant to asl for the molar concentration of 30.92% (m/v).

30.92% (m/v) means 30.92 g formalin per 100 ml of solution. If we assume that the formalin gas does not change the volume, then we have 30.92 g x 1 mol / 30.0 g = 1.031 mol / 100 ml = 10.31 mol / L = 10.31 M

2Na₃PO₄ + 3SrCl₂ ==> 6NaCl + Sr₃(PO₄)₂(s) ... balanced equation

Theoretical yield of Sr3(PO4)2(s) =

0.6735 L SrCl₂ x 0.471 mol/L x 1 mol Sr₃(PO₄)₂ / 3 mol SrCl₂ x 452.8 g Sr₃(PO₄)₂ / mol = 47.9 g Sr₃(PO₄)₂

With a 49.33% yield, the actual yield would be 0.4933 x 47.9 g = 23.6 g

NaOH(aq) + HCl(aq) ==> NaCl(aq) + H2O(l) ... balanced equation

V1M1 = V2M2

(x ml)(0.414 M) = (57.2 ml)(0.170 M)

x = 23.5 mls

Done a different way:

mols HCl present = 0.0572 L x 0.170 mol/L = 0.009724 mols

mols NaOH needed = 0.009724 since balanced eq shows us a 1:1 mol ratio between HCl and NaOH

volume NaOH needed: 0.414 mol/L (x L) = 0.009724 mols and x = 0.0235 L = 23.5 mls