Question

5. What is the osmolarity of a solution of 40 g/L Nagu (40.00 g/mole)? 6. A saline solution has an osmolarity of 0.996 OsM. What is the milliosmolarity of this solution? 7. How much 100% reagent is required to make 250 mL of 1X reagent? 8. 8. The concentration of a 1/50 dilution is 0.50 g/L. What is the starting concentration? 9. 9. How much 50 mg/dl solution is required to make 100 mL of 5 mg/dl solution? 10. 10. What is the absolute amount (grams) of substance in 5 mL of a 500 g/L solution? 11. 11. What is the maximum volume of 1 g/L dilution that can be made from 5 mL of a 500 g/L solution? 1. In the expression 8', what is the term for the numeral 8? 2. Express 1/100 in exponential form. 3. What is the specific gravity of a substance if 5 mL of the substance weighs 6.5g?

Answer 1

Molar mass of Nooh - 409 Concentration of Naon = 40g/L 402 - - 60 g/mol 4 1 molt solution IM Naot dissociates into Mat and ci in So, no. of ions produced = 2. According to definition of osmolarity Osmolarity = {Qmili. q= Osmotic cofficient or the degree of dissociation q=1 (considered as rooy, dissociation) ni = 2 no. of particles produced = 2 (Nal&ci) Cia concentration in molarity Here only one component is there Hence, Osmolarity = 4.no = 1x2x1 - 2 osmolti ... Osmolarity of 40g/ Nach is 2 osmolle.

6. Osmolarity is a S.l. unit. suli osmolarity is a ligs- unit los milanity = loo miti o smslarity 0.996 OsM = 1000 x 0.946 a DSM. - 096 morm. 7. According to equation of dilution, VC, = V₂ Cz. Via Initial wolume G, = Snitial concentration = 100x. ₂ a final volume = 250 ml Go final concentration = 1X. V,x 100= 250 ml x 1 x V Vi 250 ml x 1x -TOX - = 2.5 ml. So, 2.5 ml 100 x reagent is required.

8. According to equation of dilution : - 44 = V₂ G V, = Initial rolume of V = final volume . dilution fraction : Vašo. → 50. s= final concentration = 0.50g/l. Initial concentration, G = vzg = 50x0.50 gr = 25.9/2. Starting concentration = 25g/ .