IGCSE Chem 0620/42/F/M/2018 // Full explanation⚗// Q 1-3//Requested video//

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Check the comment written by Clara on Q-2C(IV)
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Just wanted to mention that there is a question that is solved incorrectly.
Question 2.c(iv)→When you are asked to calculate the mass of NaN03, you have to calculate the RAMS (48+23+14=85). Knowing that the relative atomic mass of sodium nitrate is 85, you have to use it in the equation (Mass =RAMs x Moles), therefore you substitute with the moles you had previously obtained from the previous question, (Mass= 85 x 4 x 10^-3), this gives a total of 0.34g.
At the start of the question it states that the yield of NaNO3 crystals was 90%, this means that you have to find 90% of the 0.34g (100% mass).
0.34 x 90/100= 0.306g.
The mass of NaN03 crystals= 0.306g

clarasanchis
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Another video thats gonna help me UR so

HayagreevaSridhar
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Thank you so much, it was very useful!

clarasanchis