chemistry
posted by amber .
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.