ALTERNATIVE PRACTICAL B D contains 6.30 g of HNO\(_3\) in 500 cm\(_3\) of solution. E was prepared by dissolving 8.20 g of washing soda crystals ( Na\(_2\)C...

Answer Details

(a) Volume of pipette used = 25.00 cm\(^3\)

Volume of A used =  \(\frac{Rough + 1st titre + 2nd titre}{3}\)

                             = \(\frac{20.90 + 20.80 + 21.00}{3}\)

                             = 20.90 cm\(^3\)

(b)  2HNO\(_3\)\(_{aq}\)  +  Na\(_2\)CO\(_3\).xH\(_2\)O\(_{aq}\) → 2NaNO\(_3\) + CO\(_2\) + (X+1)H\(_2\)0(l)

c(i) concentration of E in moldm\(^{-3}\)

Since we do not know the concentration of D in moldm\(^{-3}\), but in g per cm\(^3\), it will be difficult using the fomula \(\frac{C_A V_A}{C_B V_B}\) = \(\frac{N_A}{N_B}\).

Thus, we must first convert the Conc. of D to moldm\(^{-3}\).

We were told from the question that  D contains 6.30 g of HNO\(_3\) in 500 cm\(_3\),

So, if D            6.30 g of HNO\(_3\) was dissolved in 500 cm\(_3\)

then,                 X g of HNO\(_3\) will be dissolved in 1000 cm\(_3\)

   X =  \(\frac{{6.30}\times{1000}}{500}\) = 12.6 gdm\(^{-3}\)

Recall that  Molar concentration (moldm\(^{-3}\)) = \(\frac{Mass concentration(gdm^{-3})}{Molar mass}\)

Molar mass od HNO\(_3\) = 1 + 14 + (16 X 3) = 63 g/mol

concentration of D(moldm\(^{-3}\)) =  \(\frac{Mass concentration(gdm^{-3})}{Molar mass}\) 

                                                  C\(_D\)     =  \(\frac{12.6}{63}\) = 0.20 moldm\(^{-3}\)

Now we can use the formula \(\frac{C_D V_D}{C_E V_E}\) = \(\frac{N_D}{N_E}\).

Given that 

C\(_D\) =   0.20 moldm\(^{-3}\)                                            N\(_D\) = 2

V\(_D\) =  20.90cm\(^3\)                                                      N\(_E\) = 1

C\(_E\) = ?

V\(_E\) =  25.0cm\(^3\)

 C\(_B\) = \(\frac{C_D V_D N_E}{N_D V_E}\)

                     C\(_B\) = \(\frac{ 0.2 X 20.90 X 1}{25 X 2}\)

                                 = 0.0836 moldm\(^{-3}\) 

                                 = 0.084 moldm\(^{-3}\) 

(ii) To get the value of x in Na\(_2\)CO\(_3\).xH\(_2\)O

First, from the question, if   8.20 g of Na\(_2\)CO\(_3\).xH\(_2\)O was dissolved in 250 cm\(^3\) of distilled water.

 then,                                    Xg of Na\(_2\)CO\(_3\).xH\(_2\)O will be dissolved in 1000 cm\(^3\) of distilled water.

So,  X =  \(\frac{{8.20}\times{1000}}{250}\) = 32.8 gdm\(^{-3}\)

Mass concentration of E = 32.8 gdm\(^{-3}\)

Recall that  concentration of E(moldm\(^{-3}\)) =  \(\frac{Mass concentration(gdm^{-3})}{Molar mass}\) 

Then,  Molar mass of E ( Na\(_2\)CO\(_3\).xH\(_2\)O ) =  \(\frac{Mass concentration}{Molar concentration}\) 

                                                                                       =  \(\frac{32.8}{0.0836}\) 

                                                                                       = 392.34 g/mol

But    Molar mass of Na\(_2\)CO\(_3\).xH\(_2\)O = Na\(_2\)CO\(_3\) + x(H\(_2\)O)

                                                               392.34   = (23 x 2) + 12 + (16 x 3) + 18x

                                                               392.34  =  46 + 12 + 48 + 18x

                                                               392.34 =  106 + 18x

                                                                 18x = 392.34 - 106

                                                                  18x = 286.34

                                                                   x = \(\frac{286.34}{18}\)

                                                                   x = 15.9

                                                                   x ≈ 16