Any answer is helpful. QUESTION 1 What is the molarity of 25% w/w acetic acid so

Question

Any answer is helpful.

QUESTION 1 What is the molarity of 25% w/w acetic acid solution?(density-1.05kg/L, MW-60) Answer (mol/L) QUESTION 2 What is the molar fraction of 25% w/w acetic acid solution? (density = 1.05kg/L, MW = 60) Answer (0-1) QUESTION 3 What is the molarity of 1 % (w/v) solution of epinephrine HCl (MW-219.20)? Answer (mol/L) QUESTION 4 How many milligrams of drug X are in 10 ml of a 0.5% (w/v) solution of X? Answer (mg) QUESTION 5 What is the molecular weight of cefazolin sodium (i 2) if a 2% (w/v) aqueous solution has an osmolarity of 83.9 mOmol/L? Answer

Explanation / Answer

1) Molarity(M) = (w/w%)*d*10/Mwt

      w/w% = 25%

     d = density of solution = 1.05 kg/L = 1.05 g/ml

     Mwt = molecular weight of acetic acid solution = 60 g/mol

M = 25*1.05*10/60 = 4.375 M

2) 25%(w/w) means 25 g acetic acid present in 100 g solution.

No of mol of acetic acid = w/Mwt = 25/60 = 0.42

No of mol of water = 75/18 = 4.2 mol

molefraction of acetic acid = mols of acetic acid / total moles

                                = 0.42/(0.42+4.2)

                                = 0.091

molefraction of water = 4.2/(0.42+4.2)

                         = 0.91

3) Molarity(M) = (w/v%)*10/Mwt

      w/v% = 1%

Mwt = molecular weight = 219.2 g/mol

M = 1*10/219.2

    = 0.0456 M

4) 0.5%(w/v) means 0.5 g drug present in 100 ml solution

amount of drug X present in 10 ml solution = 0.5*10/100 = 0.05 g

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