Achemistry student must write down in her lab notebook the concentration of a solution of potassium chloride. the concentration of a solution equals the mass of what's dissolved divided by the total volume of the solution. here's how the student prepared the solution: a. the label on the graduated cylinder says: empty weight: 5.55gb. she put some solid potassium chloride into the graduated cylinder and weighed it. with the potassium chloride added, the cylinder weighed 92.7 g. c. she added water to the graduated cylinder and dissolved the potassium chloride completely. then she read the total volume of the solution from the markings on the graduated cylinder. the total volume of the solution was 23.9 ml. what concentration should the student write down in her lab notebook? be sure your answer has the correct number of significant digits.

d = 3.65 g/mL

Explanation:

This problem is solved by using the equation:

d= m/V

But we have to be careful with the number of significant figures and number of decimals to report our result.

There are two steps to calculate the density:

1. We should perform a substraction to determine the mass of potassium chloride.

mass KCl =( Weight cylinder + KCl ) - (Weight empty cylinder)

92.7 g - 5.55 g = 87.15 g = 87.2 ( rounded to 1 decimal place)

The rule for addition and substraction is that we round the result  to the number of decimal place  with the least  number of decimals ( 92.7 has one decimal, 5.55 has two)

2. We can now calculate the density by dividing the mass into the volume, but retaining the number of significant figures to the number with the smallest number of significant figures, the rule for multiplication and division.

d= m/V = 87.2 g / 23.9 mL = 3.648 g/mL

87.2 has three significant figures and so does 23.9, so we have to round to 3 significant figures.

The rule here is that if the left most digital to be dropped is greater o equal to 5, we round to the nearest higher digit, so 8 is greater than 5 and we rounded up 3.648 to 3.65.

given:

mass of na = 115 g

excess cl2

to determine:

mass of nacl produced

explanation:

given reaction is-

2na(s) + cl2(g) → 2nacl(s)

since cl2 is in excess, na will be the limiting reagent

as per the reaction stoichiometry na: nacl = 1: 1

i.e. moles of na reacted = moles of nacl formed

now, # moles of na = mass of na/atomic mass

                                  = 115 g/23 g.mol-1 = 5 moles

therefore, moles of nacl = 5

molar mass of nacl = 58 g/mol

mass of nacl = 5 moles * 58 g.mol-1 = 290 g

ans: amount of nacl produced = 290 g