# chemistry

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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

• chemistry -

massingrams=molmassKCl*.75*.29

• chemistry -

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.

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