1. What is the molarity of a solution obtained by dissolving 8.156 g of KNOs in
ID: 1028186 • Letter: 1
Question
1. What is the molarity of a solution obtained by dissolving 8.156 g of KNOs in some water and diluting to exactly 100 mL in a graduated flask? 2. How many grams of NaNO, are there in exactly 25.00 mL of a 0.1175 M aqueous solution of the salt? 3. Consider Q. 2 above. What volume of solution would you measure out to provide 0.2352 g of the salt? 4. Predict whether a precipitate will form if solutions of the following substances are mixed: a) AgNO3 +KCI b) BaCl2 + H2SO c) Pb(NO3)2 + HC1 5. For those that form a precipitate write the net ionic equation 6. If you add some vinegar (vinegar if 5% acetic acid, CH3COOH(aq)) to baking soda, a reaction takes place and a gas is released. Explain with a balanced equation.Explanation / Answer
1. Molarity(M) = (w/M)*(1000/V)
w = wt of KNO3 = 8.156 g
M = molarmass of KNO3 = 101.1 g/mol
V = vol of KNO3 solution = 100 ml
M = (8.156/101.1)*(1000/100)
= 0.807
2. nO Of mol of NaNO3 present = v*M = (25/1000)*0.1175 = 0.00294 mol
amount of NaNO3 present = n*Mwt
= 0.00294*85
= 0.25 g
3. from q.no.2 , 25 ml solution contains 0.25 g NaNO3.
so that, volume of NaNo3 containing 0.2352 g = 0.2352*25/0.25 = 23.52 ml
answer: 23.52 ml
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