How many grams of KCl are required to make .75 L of 0.29 M solution? Round to one decimal place.
i got .2907
is this correct
im not sure about the decimal places
thanks
massingrams=molmassKCl*.75*.29
Try to remember to include leading zeros as this are easily lost. So 0.75 rather than .75
MM of KCl is 74.5513 g/mol
so number of moles is
74.5513 g/mol c 0.75 litre x 0.20 mol litre^-1
=11.182 g
which rounds to 11.2 g to 1 decimal place.
You might find this site useful
http://www.cimt.plymouth.ac.uk/projects/mepres/book7/bk7i2/bk7_2i2.htm
for rounding.
To calculate the number of grams of KCl required for the solution, you need to follow these steps:
1. Start with the given information: 0.75 L of 0.29 M solution.
2. Recall that Molarity (M) is defined as the number of moles of solute per liter of solution (mol/L). In this case, the molarity is 0.29 M.
3. Convert the molarity to moles per liter (mol/L). This means that there are 0.29 moles of KCl in every liter of the solution.
4. Multiply the molarity by the volume of the solution in liters to find the number of moles of KCl needed. So, 0.29 mol/L x 0.75 L = 0.2175 moles of KCl.
5. Lastly, calculate the number of grams of KCl by multiplying the number of moles by the molar mass of KCl. The molar mass of KCl is 74.55 g/mol (39.10 g/mol for K + 35.45 g/mol for Cl). So, 0.2175 mol x 74.55 g/mol = 16.1729 grams.
Rounding to one decimal place, the answer is approximately 16.2 grams of KCl needed to make 0.75 L of 0.29 M KCl solution.
So, while your value of 0.2907 is very close, it is not rounded to one decimal place. The correct answer, rounded to one decimal place, is 16.2 grams.