[Sis290025]

Please tell me if my calculations are correct and if I expressed the answer with the correct number of significant figures.

What will be the mass of antifreeze in tons added to a car when the radiator is filled with antifreeze (d = 1.12 g/mL) if the radiator has a capacity of 6.84 gallons?

m/V = 1.12 g/mL

m = d*V

V = 6.84 gal*(3.7854 L /1 gal)*(1000 mL/1L) = 25892.136 mL

m_antifreeze = (1.12 g/mL)* (25892.136 mL)*(1 kg / 1000 g)*(1 ton / 1000 kg) = 2.90*10^-2 tons of antifreeze

[sjb]

Please tell me if my calculations are correct and if I expressed the answer with the correct number of significant figures.

What will be the mass of antifreeze in tons added to a car when the radiator is filled with antifreeze (d = 1.12 g/mL) if the radiator has a capacity of 6.84 gallons?

m/V = 1.12 g/mL

m = d*V

V = 6.84 gal*(3.7854 L /1 gal)*(1000 mL/1L) = 25892.136 mL

m_antifreeze = (1.12 g/mL)* (25892.136 mL)*(1 kg / 1000 g)*(1 ton / 1000 kg) = 2.90*10^-2 tons of antifreeze

I disagree, personally, if only to be pedantic and say that for me a gallon is not 3.7854 litres. See e.g. http://en.wikipedia.org/w/index.php?title=Gallon&oldid=280878132 for various definitions.

[Sis290025]

Thank you very much for your input.

[sjb]

That said, if you've used an appropriate gallon, then the maths looks right.