Molarity (without stoich)
Quiz
•
Science
•
10th Grade
•
Medium
Standards-aligned
Samina syed
Used 2+ times
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13 questions
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1.
MULTIPLE CHOICE QUESTION
5 mins • 1 pt
How many liters would you need to make a 1 M solution if you have 6 mol of Sodium Hydroxide?
2
3
4
6
Answer explanation
To make a 1 M solution with 6 mol of Sodium Hydroxide, you would need 6 liters of the solution.
2.
MULTIPLE CHOICE QUESTION
5 mins • 1 pt
What is the molarity of 4 g of NaCl (MM=58.45) in 3,800 mL of solution?
0.018 M
0.0011 M
1.052 M
0.062 M
Answer explanation
To find the molarity, first calculate the moles of NaCl: 4g / 58.45 g/mol = 0.0684 mol. Then, divide moles by liters: 0.0684 mol / 3.8 L = 0.018 M.
3.
MULTIPLE CHOICE QUESTION
45 sec • 1 pt
Find the molarity of 186.55 g of sucrose, C12H22O11 (MM = 342) in 250 mL of water.
2.18 M
0.746 M
1.18 M
0.545 M
Answer explanation
The molarity is calculated by dividing the moles of sucrose by the volume in liters. Moles = 186.55g / 342g/mol = 0.545 mol. Molarity = 0.545 mol / 0.25 L = 2.18 M.
4.
MULTIPLE CHOICE QUESTION
1 min • 1 pt
How many grams of AgNO3 (MM = 169.87) are needed to prepare 0.125M solution in 250 mL of water?
.03g
0.5g
5.3g
84.9g
Answer explanation
To calculate the grams needed, use the formula: grams = Molarity x Volume x Molar Mass. Substituting the values, grams = 0.125 x 0.25 x 169.87 = 5.3g. Therefore, 5.3g of AgNO3 are needed to prepare the solution.
5.
MULTIPLE CHOICE QUESTION
1 min • 1 pt
Calculate the molarity of the following solution: 1.0 mole of KCl in 750.0 mL of solution.
0.750 M
99 M
1.3 M
2.0 M
Answer explanation
The molarity is calculated by dividing moles of solute by liters of solution. In this case, 1.0 mole of KCl in 750.0 mL (0.75 L) gives a molarity of 1.3 M
6.
MULTIPLE CHOICE QUESTION
3 mins • 1 pt
What is the molarity of a solution made by adding 1.565 moles of PbNO3 to 500 mL?
300. M
31.3 M
3.13 M
1.56 M
Answer explanation
To find the molarity, divide the moles of solute by the volume of solution in liters. 1.565 moles / 0.5 L = 3.13 M.
7.
MULTIPLE CHOICE QUESTION
5 mins • 1 pt
How many L are required to make 3.5 M hydrochloric acid using 1.1 moles?
3.18 L
0.31 L
3.85 L
4.6 L
Answer explanation
To make 3.5 M HCl with 1.1 moles, divide moles by concentration: 1.1 moles / 3.5 M = 0.31 L required.
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