Question

Question 23 (1 point) A student runs the reaction below with 201 mL of 0.501 M sulfuric acid (H2SO4). If the sodium hydroxide is 0.280 M, how many milliliters will be needed for the reaction? H2SO4 + 2 NaOH Na2SO4 + 2 H20 Report your answer in milliliters (mL) with the correct number of significant figures. Your Answer: Answer units Question 24 (1 point) How many moles of sodium chloride would be contained in 820.0 mL of a 0.41 M solution? E o i 8 N 5 6 P Y U R К. G F H

Answer 1

Chemical Calculation Suggested solution no. 23 Given:- Volume of sulfuric acid (H2SO4) = 201 mL Molarity of sulfuric acid (H2SO4) = 0.501 M Molarity of sodium hydroxide (NaOH) = 0.280 M For neutralization, reaction mili-equivalents of acid (meq)a should be equal mili-equivalent base (meq)b- (meq)a = (meq) ............(i) mili-equivalent of acid = Normality of acid(N) x volume of acid in mL Normality (N) = molarity (M) ~ n-factor For sulfuric acid (H2SO4) n-factor = 2.0 (2H* replaceable) For NaOH n-factor is = 1.0 (one -OH replaceable) Mili-equivalent of acid (meq)a Mili-equivalent of base (meq) H2SO4 NaOH Normality (N) < Volume in mL N (Normality) Volume in mL Molarity (M) xn- factor x Molarity (M) xn- factor x volume ml volume ml 0.501 M X 2.0 X 201 mL 0.280 MX 1.0 x Vol(mL) Now apply equation no.(i), we get (meq), H2SO4 = (meq)NaOH...... (i) 0.501 M X 2.0 201 mL = 0.280 M X Vol (ml) 0.501 M X2.0X201 mL Volume of NaOH (mL) = 0.280 M Answer should be in two significant figure Volume of NaOH = 719.3 mL = 719 mL volume of NaOH = 720 ml

Suggested solution no-24 (i) weight in gram 1000 Molarity (M) Х molecular weight volume in ml Rewrite equation molarity (M)xmolecular weight xvolume in ml weight in gram = 1000 Given :- Molarity of NaCl = 0.41 M Volume of solution (VmL) = 820 mL Molecular weight of NaCl = 23 + 35.5 = 58.5 gram-mol Now apply equation (ii), we get 0.41 MX58.5 gram-mole-1x820 ml weight in gram = = 19.6677 =19.67 gram 1000