I believe a is right with b and c wrong.
For d you appear to have rounded the answer to 15g; however, you ha e three significant figures is 35.0 and 4 s.f. in 12.00M, so don't throw a good digit away. You should have 3 s.f. in your answer.
b. mols KOH = 3.5/56.1 = ?mols
M = ?mols/0.150 = ? about 0.4M
c. mols KOH = 5.00/56.1 = about 0.89
M = mols/L or L = mols/M = 0.089/0.400M = about 0.22L
My answers are close; you should redo them for a more accurate answer. Also I've not kept up with the s.f. in my answers since they are estimates.
a.)How many moles of NaOH are required to prepare 2.00L od 0.380M solution? answer is 0.760 mole of NaOH (mole/l * 2.00L)
b.)What will be the molarity of a solution if 3.50 g of KOH are dissolved in water to make 150.0 mL of solution? answer is 0.52 M KOH (gKOH/molar mass KOH=mole KOH; M=mol KOH/L)
c.)What volume of (mL) of 0.400M solution can be prepared by dissolving 5.00g of KOH in water? answer is 300mL (gKOH/molar mass=mole KOH;mole KOH * Molarity of the solution 0.400M=L of solution *ml)
d.) Calculate the weight of HCl in 35.0mL of comcentrated HCl (12.00M) solution. my asnswer is 15 g HCl ( mL to L; L * M=mol; mole* molar mass= g)
pls check my answers.
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