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Help me to find out Density problem.


SophiaRivera007
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The density of two liquids (A and B) is given as 1000 kg/m3 and 600 kg/m3, respectively. The two liquids are mixed in a certain proportion and the density of the resulting liquid is 850 kg/m3. How much of liquid B (in grams) does 1 kg of the mixture contain? Assume the volume of the two liquids is additive when mixed.

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Moderator Note

Since my last note was an unofficial probe as a moderator, I'm just going to come back and make this clear.

This is a homework HELP forum, not a we'll do your homework for you forum. I had thought I had made that abundantly clear in my initial post, but apparently not. The OP was made not even a day ago. They might not have had a chance to log back on yet. Whatever the case, it would be greatly appreciated if you could all refrain from just giving out answers. Doing so helps absolutely no one, least of all the OP.

I have hidden the offending posts.

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First, let VA represent the volume of liquid A, and let VB represent the volume of liquid B, in 1 kg of mixture. Then, 850x(VA+VB) = 1 kg (of mixture). In addition, 1000VA+600VB = 1 kg (of mixture). We have here two equations with two unknowns (VA and VB). Solve for VB. The mass of liquid B in 1 kg of mixture is 600VB, which is equal to 264.7 grams.

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First, let VA represent the volume of liquid A, and let VB represent the volume of liquid B, in 1 kg of mixture. Then, 850x(VA+VB) = 1 kg (of mixture). In addition, 1000VA+600VB = 1 kg (of mixture). We have here two equations with two unknowns (VA and VB). Solve for VB. The mass of liquid B in 1 kg of mixture is 600VB, which is equal to 264.7 grams.

 

Thank you AshBox. I find the perfect solution.

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First, let VA represent the volume of liquid A, and let VB represent the volume of liquid B, in 1 kg of mixture. Then, 850x(VA+VB) = 1 kg (of mixture). In addition, 1000VA+600VB = 1 kg (of mixture). We have here two equations with two unknowns (VA and VB). Solve for VB. The mass of liquid B in 1 kg of mixture is 600VB, which is equal to 264.7 grams.

Doesn't a modnote mean anything?

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