Answer to Question #168085 in General Chemistry for Dak Ruot Tiptip

Question

(1) A laboratory experiment calls for 0.125M of HNO3. What volume of 0.125 M of HNO3 can be prepared from 0.250L of 1.88 g/m of HNO3?

(2) calculate the molarity of 1.49kg of isopropyl alcohol, C3H7OH, in 2.50L of solution, the concentration of isopropyl alcohol in rubbing alcohol.

(3) In canada and united kingdom, devices that measure blood glucose levels provide a reading in mM per litre. If a measurement of 5.3 mM is observed, what is the concentration of glucose?

(3) what mass of solid NAOH (97% NAOH by mass) is required to prepare 1L of a 10% solution of NAOH by mass? The density of the 10% solution is 1.09g/m

Answer 1

(1) m(HNO3) = (250 mL)*(1.88 g/mL) = 470 g

n(HNO3) = (470 g)/(63 g/mol) = 7.46 mol

V = (7.46 mol)/(0.125M) = 59.68 L.

(2) n(C3H7OH) = (1490 g)/(60 g/ mol) = 24.8333 mol

C = (24.8333 mol)/(2.50 L) = 9.93M.

(3) Glucose concentration is usually measured in mg/dL. 1 L = 10 dL.

C(C6H12O6) = (5.3 mM)*(1 L)*(180 g/mol)/(10 dL) = 95.4 mg/dL

(4) m(solution) = (1.09 g/mL)*(1000 mL) = 1090 g

m(NaOH) = 0.1*(1090 g) = 109 g

m(NaOH solid) = (109 g)/0.97 = 112.37 g