Asked 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
i got .2907
is this correct
im not sure about the decimal places
thanks
Answers
Answered by
bobpursley
massingrams=molmassKCl*.75*.29
Answered by
Dr Russ
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.
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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