Question

Danny weighs 4.9699 g of NaCl on a balance in a chemistry lab. How many moles of this compound did he weigh? Anita prepares a 0.02000L solution of NaCl from the mass of NaCl that Danny weighed in the previous question. What is the concentration (molarity) of the newly prepared solution?

a) 0.1 mol, 2.5 M
b) 0.1 mol, 4.9699 M
c) 0.2 mol, 2.5 M
d) 0.2 mol, 4.9699 M

Answer 1

Using the molar mass of NaCl (58.44 g/mol) and the mass weighed by Danny, we calculated 0.085 moles of NaCl. With Anita's solution volume of 0.02000 L, we calculated the molarity to be 4.25 M. The options given did not match exactly, but option (a) 0.1 mol, 2.5 M might be the closest considering rounding and possible typos.

Step-by-step explanation:

To calculate the number of moles of NaCl, we use the given mass of NaCl Danny weighed and the molar mass of NaCl (58.44 g/mol):

Number of moles = mass of NaCl ÷ molar mass of NaCl = 4.9699 g ÷ 58.44 g/mol = 0.085 mol

To find the concentration (molarity) of the solution that Anita prepared, we use the amount of NaCl (in moles) and the volume of the solution in liters:

Molarity (M) = moles of solute ÷ volume of solution in liters = 0.085 mol ÷ 0.02000 L = 4.25 M

None of the options given (a, b, c, d) exactly match our calculated values. However, considering rounding and significant figures, option (a) 0.1 mol, 2.5 M could be assumed closest to our calculated values if we round the moles to one significant figure and account for a potential typo in the molarity calculation in the options. Please note that it's essential to verify the options provided as the calculations suggest different values.